Prove that the distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely 1: 3: 5: …………….].

Options

A. `1:3:5:7:9:`....
B. `2:4:6:8:10:`....
C. `1:4:7:8:10:`.......
D. None of these

Answer

A

Explanation

Here, u=0, a=g
Distance travelled in `n^(th)` second is given by
`D_(n)=u+(a)/(2)(2n-1)`
Distance travelled in `1^(st)` second (i.e. when n=1) will be
`D_(1)=0+(g)/(2)(2xx1-1)=(g)/(2)`
Distance travelled in `3^(rd)` second (i.e. when (i.e. when n=3)
will be
`D_(3)=0+(g)/(2)(2xx3-1)=(5g)/(2)`
Distance travelled in `4^(th)` second (i.e. when n=4)
will be
`D_(4)=0+(g)/(2)(2xx4-1)=(7g)/(2)`
Distance travelled in `5^(th)` second (i.e. when n=5)
will be
`D_(5)=0+(g)/(2)(2xx5-1)=(9g)/(2)`
`therefore D_(1):D_(2):D_(3):D_(4):D_(5)`..........=1:3:5:7:9:.........